2020 AMC 12B Problems/Problem 6
Problem
For all integers 
the value of
![\[\frac{(n+2)!-(n+1)!}{n!}\]](http://latex.artofproblemsolving.com/e/4/c/e4cf3e216ff29ec12bb85e389993280b1ca40180.png)
is always which of the following?
Solution 1
We first expand the expression:
![\[\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}\]](http://latex.artofproblemsolving.com/f/b/b/fbb7fd2ba8842e956cbb39cd25e8f8e1afcb1f1b.png)
We can now divide out a common factor of 
from each term of this expression:
![\[(n+2)(n+1)-(n+1)\]](http://latex.artofproblemsolving.com/a/8/a/a8a064934f7783af35d33f76ddd140390f109856.png)
Factoring out 
we get ![\[(n+1)(n+2-1) = (n+1)^2\]](http://latex.artofproblemsolving.com/2/9/c/29cf1b77a7720e7818dcafab9161cab373f29495.png)
which proves that the answer is 
Solution 2
In the numerator, we factor out an to get ![\[\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).\]](http://latex.artofproblemsolving.com/6/6/0/6605f384fc7eb446d310e1fc0b187538f98f32c1.png)
Now, without loss of generality, test values of 
until only one answer choice is left valid:
knocking out 
and 
knocking out 
and 
knocking out 
knocking out 
This leaves 
as the only answer choice left.
This solution does not consider the condition 
The reason is that, with further testing it becomes clear that for all 
we get ![\[(n+2)(n+1)-(n+1) = (n+1)^{2},\]](http://latex.artofproblemsolving.com/2/2/7/227c9dcc1a25f21fe49c32d711cf9a9fb6086260.png)
as proved in Solution 1. The condition 
was most likely added to encourage the selection of choice 
~DBlack2021 (Solution Writing)
~Countmath1 (Minor edits in formatting)
~MRENTHUSIASM (Edits in Logic)
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2234
~ pi_is_3.14
Video Solution
~IceMatrix
See Also
| 2020 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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